高等数学积分问题
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√(2x-x^2) = √[1-(1-x)^2],
令 1-x = sint， 则 x = 1-sint， dx = -costdt
V = 2π ∫<π/2, 0> (1+sint) cost(-cost)dt
= 2π ∫<0, π/2> (1+sint) (cost)^2dt
= 2π ∫<0, π/2> (cost)^2dt + 2π ∫<0, π/2> sint(cost)^2dt
= π ∫<0, π/2> (1+cos2t)dt - 2π ∫<0, π/2> (cost)^2dcost
= π [t+(1/2)sin2t]<0, π/2> - (2π/3)[(cost)^3]<0, π/2>
= π^2/2 + 2π/3
令 1-x = sint， 则 x = 1-sint， dx = -costdt
V = 2π ∫<π/2, 0> (1+sint) cost(-cost)dt
= 2π ∫<0, π/2> (1+sint) (cost)^2dt
= 2π ∫<0, π/2> (cost)^2dt + 2π ∫<0, π/2> sint(cost)^2dt
= π ∫<0, π/2> (1+cos2t)dt - 2π ∫<0, π/2> (cost)^2dcost
= π [t+(1/2)sin2t]<0, π/2> - (2π/3)[(cost)^3]<0, π/2>
= π^2/2 + 2π/3
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